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jek_recluse [69]
3 years ago
6

HOW DOES DENSITY AFFECT THE MELTING POINT OF WATER?

Chemistry
1 answer:
ddd [48]3 years ago
5 0
<span> As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. Bubbles of vaporized liquid (i.e., gas) form within the bulk liquid and then rise to the surface where they burst and release the gas. (At the boiling temperature the vapor inside a bubble has enough pressure to keep the bubble from collapsing.) In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them.<span> The temperature of a boiling liquid remains constant, even when more heat is add.</span></span>
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In a titration to find the concentration of 30ml of a H2SO4 solution, a student found that 40ml of 0.2M KOH solution was needed
lianna [129]

the queation is so difficult

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2 years ago
How do I "predict the colour change that would be observed if carbon dioxide gas were bubbled into a mixture of calcium oxide, b
Effectus [21]
Hey there!

Just say what color you think the mixture would look like if those elements were combined. Personally I don't know because I don't have context, but if it comes to it just pick a color :)

Good luck, have a good night.
7 0
3 years ago
Which graph shows the relationship between the temperature and volume of agas according to Charles's law?Volume vs. Temperature
solmaris [256]
Graph D.

Charles's law states that at the same pressure, the Volumen of a gas is proportional to its temperature (absolute temperature).

This is

V = K T, where K is the constant of proportionallity.

You can also write it in the form

V1 / V2 = T1/ Ts

When Temperature rise Volumen rise in a proportional way. If Temperature doubles Volumen doubles, if Tempearature is decreases to on third, Volume will decrease to one third.
5 0
3 years ago
At 25 °c and 785 torr, carbon dioxide has a solubility of 0.0351 m in water. what is its solubility at 25 °c and 1510 torr?
jonny [76]
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4 0
3 years ago
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
5 0
3 years ago
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